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How to get one from many?

  1. Apr 30, 2008 #1
    First of all, since my title isn't self explanatory, what I am trying to do is take a large number and work with it such that overtime it becomes a small number. For example, I am trying to take a number like 400 and say that it came from the number one. But it came from the number 1 in such a way that every 20 hours, the number 1 duplicated creating two. 20 hours later again those two duplicated creating four, another 20 hours and 4 makes 8. Is there a way to do this easily backwards from 400 and see how long it takes to get to one? Also, to throw in a twist, is there a way to figure out the same thing, but every 24 hours one out of the bunch is lost. Like say a Parents having kids, there are 400 kids today and we want to figure out how long it took to make those 400 kids from two people, including the death along the way? Is this possible using math on paper, or would it require something more?

  2. jcsd
  3. Apr 30, 2008 #2


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    That's the binary (base-2) logarithm: [itex]\log_2400\approx8.64[/itex].

    It can be done, but you'd need to specify how the death works. For example, if people die on the third generation like so:


    then this can be described mathematically by the recurrence relation [itex]n_k=2n_{k-1}-n_{k-3}[/itex] for [itex]k\ge4[/itex] with starting condition [itex]n_1=2,n_2=4,n_3=8[/itex].
  4. Apr 30, 2008 #3
    we get the weirdest people on these boards
  5. May 1, 2008 #4
    Are you looking for a formula that includes EXACTLY 400 at some point?
  6. May 1, 2008 #5

    This depends on how you want your numbers to work out:

    If you want "births" and "deaths" to be continuous (for instance, with death rate = 0 if right now there is 1 person and in 2 minutes there will be 2 people, then you know that in 1 minute, there will be Sqrt[2] people)
    then you can write a differential equation describing the situation
    in particular, for the case where it's just people being born and people dying and the birth rate and death rates are both proportional to the population, this is just
    f'(t) = (b - d) f(t)
    (where b and d are the birth and death rates per population)

    The solution to this is f(t) = e^[(b-d)t]

    This solution has the perhaps undesirable result that even 1/2 a person will contribute to producing another person (and that half a person can die and that after half a person dies, the other people will die a little slower)

    However, the theory of ordinary differential equations is fairly well developed, and if you have some Calculus knowledge, you'll probably get the hang of writing and solving some of the easier ones eventually. (Some of the harder ones on the other hand can be extremely difficult to solve. It can be shown that many differentially equations are in fact impossible to solve with a simple solution)


    If on the other hand you want your births and deaths to come in discrete amounts so that it doesn't make any sense to say that 1/2 a person was born or died, then you can describe this with a recurrence relation such as the one CRGreathouse described.

    Again, there is a theory of recurrence relations that has been developed to solve them, but they are generally more difficult than differential equations. Often when dealing with very large numbers, people ignore the discreteness of a system and pretend that it is continuous so that they can write down a differential equation that might be easier to solve.


    As a side note, one thing that I find quite interesting is that often when you try to solve more difficult difficult differential equations, as part of the solution, you will need to solve a recurrence relation. Likewise, solving recurrence relations often involves solving a differential equation. So if you're interested in solving problems of the sort you're asking about, it'd be useful to learn how to solve both types.

    Differential Equations is considered its own subject, so once you know Calculus, you can learn to solve differential equations by reading books on the subject (what I mean is that this is the title of the subject you want to look up). On the other hand, recurrence relations are usually considered part of the larger problem of counting discrete sets of objects in general (since they are used in and make use of many techniques for counting); the title of this subject is Combinatorics.
    Last edited: May 1, 2008
  7. May 1, 2008 #6
    In all honesty, what I am trying to do is track back my cells. From the number of cells that I have today, back to the original gametes. I'm trying to determine how many cells I have gone through, and how long it took for me to go through them mathematically. Obviously I can add roughly 9 months to my age, but I want to do it mathematically, as well as see how many cells i have gone through. I will look up those two subjects, from what you are all saying (except for guy calling me weird) then I might just be able to handle this level of math. I am high school junior, but I have some calculus knowledge (I don't know if some is enough, but I am taking IB maths, so...I don't know). It sounds like a fun endeavor. I think my next step is to track how many cells not only I, but my ancestors before me were used up and the time it took them to die. I'll go back as far as I know my family history to, then I maybe I can theoretically go back to mitochondrial eve.
  8. May 2, 2008 #7
    This will be quite the difficult task indeed!

    Your best bet is to start by making some pretty large assumptions to build your model, and then refining them so that you get a better and better aproximation.

    You will also need some vast knowledge of human cells and should break down your boddy into major tissue groups. This is because certain cells (belonging to different types of tissue) in your body will live longer than others. For example, from what I found on a different forum, skin cells live for about 1 month, while brain cells live for about 50 years. You will aslo need to know the rate at which these tissue groups grow.

    You can then use the rule that CRHouse to begin your approximation.
  9. May 3, 2008 #8


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    If we know that the number doubles every 20 hours, then we know that 20 hours ago it was half of what it is now. 40 hours ago it was 1/4 of what it is now; 60 hours ago it was 1/8 of what it is now, etc. etc.

    If you're looking for an equation:

    2^{(H/20)} = 400

    where H is the number of hours since the starting point (when the number was at 1).

    Take the log of both sides of the equation:

    H/20 = \log(400) / \log(2)

    H = 20 \cdot \log(400) / \log(2)

    So there have been [tex]\log(400) / \log(2) = 8.64[/tex] doublings to get from 1 to 400, which at 20 hours per doubling means 173 hours.

    If you want to use numbers other than 20 hours and 400, just replace the 20 and the 400 in the above equation to see what you get.

    Hope that helps.
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